document.write( "Question 1197652: The growth model for a population of bacteria is given by P(t)=35e^2t where t is measured in days. How many days will it take until the number of bacteria has exceeded 17,500? \n" ); document.write( "
Algebra.Com's Answer #831012 by math_tutor2020(3816)\"\" \"About 
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\n" ); document.write( "A useful saying: \"If the variable is in the trees, then log it down\". By \"trees\", I'm referring to exponents.\r
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\n" ); document.write( "\n" ); document.write( "We'll use logarithms to isolate the exponent in the variable, so we can isolate the variable itself.\r
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\n" ); document.write( "\n" ); document.write( "With that in mind, we replace P(t) with 17500 and solve like so.
\n" ); document.write( "P(t) = 35e^(2t)
\n" ); document.write( "17500 = 35e^(2t)
\n" ); document.write( "17500/35 = e^(2t)
\n" ); document.write( "500 = e^(2t)
\n" ); document.write( "Ln(500) = Ln( e^(2t) ) ... this is where logs come in
\n" ); document.write( "Ln(500) = 2t*Ln(e) .... use the rule log(A^B) = B*log(A)
\n" ); document.write( "Ln(500) = 2t*1
\n" ); document.write( "2t = Ln(500)
\n" ); document.write( "t = Ln(500)/2
\n" ); document.write( "t = 3.1073040492111
\n" ); document.write( "It will take a little over 3 days to reach a population of 17,500 bacteria.\r
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\n" ); document.write( "\n" ); document.write( "For day 0, aka the starting day, we have:
\n" ); document.write( "P(t) = 35e^(2t)
\n" ); document.write( "P(0) = 35e^(2*0)
\n" ); document.write( "P(0) = 35e^(0)
\n" ); document.write( "P(0) = 35*1
\n" ); document.write( "P(0) = 35
\n" ); document.write( "There are 35 bacteria at the start.\r
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\n" ); document.write( "\n" ); document.write( "Let's see how many there are for day 1.
\n" ); document.write( "P(t) = 35e^(2t)
\n" ); document.write( "P(1) = 35e^(2*1)
\n" ); document.write( "P(1) = 35e^(2)
\n" ); document.write( "P(1) = 35*7.38905609893584
\n" ); document.write( "P(1) = 258.616963462754
\n" ); document.write( "So about 258 bacteria if we only consider whole numbers of bacteria.\r
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\n" ); document.write( "\n" ); document.write( "Now day 3
\n" ); document.write( "P(t) = 35e^(2t)
\n" ); document.write( "P(3) = 35e^(2*3)
\n" ); document.write( "P(3) = 35e^(6)
\n" ); document.write( "P(3) = 35*403.428793493586
\n" ); document.write( "P(3) = 14,120.0077722756
\n" ); document.write( "About 14120 bacteria at this point. We're getting pretty close to the target 17,500.\r
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\n" ); document.write( "\n" ); document.write( "Let's look at day 4
\n" ); document.write( "P(t) = 35e^(2t)
\n" ); document.write( "P(4) = 35e^(2*4)
\n" ); document.write( "P(4) = 35e^(8)
\n" ); document.write( "P(4) = 35*2980.9579870501
\n" ); document.write( "P(4) = 104,333.529546753
\n" ); document.write( "We've reached about 104333 bacteria which is definitely way over the target of 17,500.\r
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\n" ); document.write( "\n" ); document.write( "These numeric results help show that we reach a population of 17,500 in the interval 3 < t < 4, which helps confirm the answer we reached a while back earlier.\r
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\n" ); document.write( "\n" ); document.write( "Answer: Approximately 3.107304 days.
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